A diver enters the water with an initial velocity of $28.2 \, m/s$ and comes to a stop after traveling a distance of $4.00 \, m$. Assuming uniform acceleration, we need to: a) Calculate the average acceleration of the diver. b) Calculate the time it takes for the diver to come to a stop. c) Calculate the speed of the diver after traveling $2.00 \, m$.

1. Problem Description

A diver enters the water with an initial velocity of $28.2 \, m/s$ and comes to a stop after traveling a distance of $4.00 \, m$. Assuming uniform acceleration, we need to:

a) Calculate the average acceleration of the diver.

b) Calculate the time it takes for the diver to come to a stop.

c) Calculate the speed of the diver after traveling $2.00 \, m$.

2. Solution Steps

a) Calculate the average acceleration:

We know the initial velocity $v_i = 28.2 \, m/s$, the final velocity $v_f = 0 \, m/s$, and the distance $\Delta x = 4.00 \, m$. We can use the following kinematic equation to find the acceleration $a$:

$v_f^2 = v_i^2 + 2 a \Delta x$

$0^2 = (28.2)^2 + 2 \cdot a \cdot 4.00$

$0 = 795.24 + 8a$

$a = -\frac{795.24}{8}$

$a = -99.405 \, m/s^2$

b) Calculate the time to stop:

We can use the following kinematic equation to find the time $t$:

$v_f = v_i + at$

$0 = 28.2 + (-99.405)t$

$t = \frac{-28.2}{-99.405}$

$t = 0.2837 \, s$

c) Calculate the speed after 2.00 m:

We can use the kinematic equation again with $\Delta x = 2.00 \, m$:

$v_f^2 = v_i^2 + 2 a \Delta x$

$v_f^2 = (28.2)^2 + 2 \cdot (-99.405) \cdot 2.00$

$v_f^2 = 795.24 - 397.62$

$v_f^2 = 397.62$

$v_f = \sqrt{397.62}$

$v_f = 19.94 \, m/s$

3. Final Answer

a) The average acceleration of the diver is $-99.405 \, m/s^2$.

b) The time it takes for the diver to come to a stop is $0.2837 \, s$.

c) The speed of the diver after diving $2.00 \, m$ is $19.94 \, m/s$.